import java.util.LinkedList;
import java.util.Queue;

public class BinaryTree {


    static class TreeNode {
        public char val;
        public TreeNode left;//左孩子的引用
        public TreeNode right;//右孩子的引用

        public TreeNode(char val) {
            this.val = val;
        }
    }


    /**
     * 创建一棵二叉树 返回这棵树的根节点
     *
     * @return
     */
    public TreeNode createTree() {
        TreeNode A=new TreeNode('A');
        TreeNode B=new TreeNode('B');
        TreeNode C=new TreeNode('C');
        TreeNode D=new TreeNode('D');
        TreeNode E=new TreeNode('E');
        TreeNode F=new TreeNode('F');
        TreeNode G=new TreeNode('G');
        //构建起节点之间的联系
        A.left=B;
        A.right=C;
        B.left=D;
        B.right=E;
        C.left=F;
        C.right=G;
        return A;
    }

    // 前序遍历
    public void preOrder(TreeNode root) {
        if (root==null){
            return;
        }
        System.out.print(root.val+" ");
        preOrder(root.left);
        preOrder(root.right);
    }

    // 中序遍历
    void inOrder(TreeNode root) {
        if (root==null){
            return;
        }
        inOrder(root.left);
        System.out.print(root.val+" ");
        inOrder(root.right);
    }

    // 后序遍历
    void postOrder(TreeNode root) {
        if (root==null){
            return;
        }
        postOrder(root.left);
        postOrder(root.right);
        System.out.print(root.val+" ");
    }

    public static int nodeSize=0;

    /**
     * 获取树中节点的个数：遍历思路
     */
    void size(TreeNode root) {
        if (root==null){
            return ;
        }
        nodeSize++;
        size(root.left);
        size(root.right);
    }

    /**
     * 获取节点的个数：子问题的思路
     *
     * @param root
     * @return
     */
    int size2(TreeNode root) {
        if (root==null){
            return 0;
        }
        return size2(root.left)+size2(root.right)+1;
    }


    /*
     获取叶子节点的个数：遍历思路
     */
    public static int leafSize = 0;

    void getLeafNodeCount1(TreeNode root) {
        if (root==null){
            return;
        }
        //判断当前root结点是不是叶子节点
        if (root.left==null&&root.right==null){
          leafSize++;
        }
        getLeafNodeCount1(root.left);
        getLeafNodeCount1(root.right);

    }

    /*
     获取叶子节点的个数：子问题
     */
    int getLeafNodeCount2(TreeNode root) {
        if (root==null){
            return 0;
        }
        if (root.left==null&&root.right==null){
            return 1;
        }
        return  getLeafNodeCount2(root.left)+ getLeafNodeCount2(root.right);
    }

    /*
    获取第K层节点的个数
     */
    int getKLevelNodeCount(TreeNode root, int k) {
        if (root==null){
            return 0;
        }
        if(k==1){
            return 1;
        }

        return getKLevelNodeCount(root.left,k-1)+getKLevelNodeCount(root.right,k-1);
    }

    /*
     获取二叉树的高度
     时间复杂度：O(N)
     */
    int getHeight(TreeNode root) {
          if (root==null){
              return 0;
          }
          int leftH=getHeight(root.left);
          int rightH=getHeight(root.right);
          return leftH>rightH ? leftH+1:rightH+1;
    }


    // 检测值为value的元素是否存在
    TreeNode find(TreeNode root, char val) {
        if (root==null){
            return null;
        }
        //判断当前root值，是不是我要找的value
        if (root.val==val){
            return root;
        }
      //递归root的左子树和右子树，去左子树和右子树里面去找
        TreeNode retL=find(root.left,val);
        if (retL!=null){
            return retL;
        }
        TreeNode retR=find(root.right,val);
        if (retR!=null){
            return retR;
        }
        return null;
    }

    //层序遍历
    void levelOrder(TreeNode root) {
   if (root==null){
       return;
   }
   //借助队列，完成层序遍历
        Queue<TreeNode>  queue=new LinkedList<>();
      queue.offer(root);
      while (!queue.isEmpty()){
          TreeNode cur=queue.poll();
          System.out.print(cur.val+" ");
          if (cur.left!=null){
              queue.offer(cur.left);
          }
          if (cur.right!=null){
              queue.offer(cur.right);
          }
      }
    }


    // 判断一棵树是不是完全二叉树
    boolean isCompleteTree(TreeNode root) {
        if (root==null){
            return true;
        }
        //借助队列，完成层序遍历
        Queue<TreeNode>  queue=new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()){
            TreeNode cur=queue.poll();
            if (cur!=null){
                queue.offer(cur.left);
                queue.offer(cur.right);
            }
            else {
                break;
            }
             }
        while (!queue.isEmpty()) {
            TreeNode cur=queue.poll();
            if (cur!=null){
                return false;
            }
        }
        return true;
    }

}